*(Then t-x = 0 which means the integral can be further simplified) According to my understanding, t is the time domain and any domain x can be ...*

It is instead done so that we can note that we did this integral back in the Fourier sine series section and so don’t need to redo it in this section.

Using the previous result we get, \[ = \frac\hspace\hspacen = 1,2,3, \ldots \] In this case the Fourier series is, \[f\left( x \right) = \sum\limits_^\infty \sum\limits_^\infty = \sum\limits_^\infty \] If you go back and take a look at Example 1 in the Fourier sine series section, the same example we used to get the integral out of, you will see that in that example we were finding the Fourier sine series for \(f\left( x \right) = x\) on \( - L \le x \le L\).

\[\begin & = \frac\int_^ = \frac\left[ \right]\\ & = \frac\left[ \right] = \frac\left[ \right] = L\end\] \[\begin = \frac\int_^ & = \frac\left[ \right]\\ & = \frac\left[ \right]\end\] At this point it will probably be easier to do each of these individually. \right|_^0 = \frac\sin \left( \right) = 0\] \[\begin\int_^ & = \left.

\right|_0^L\\ & = \left( \right)\left( \right)\\ & = \left( \right)\left( \right)\end\] So, if we put all of this together we have, \[\begin & = \frac\int_^ = \frac\left[ \right]\\ & = \frac\left( \right)\,\,\,,\hspace\hspacen = 1,2,3, \ldots \end\] So, we’ve gotten the coefficients for the cosines taken care of and now we need to take care of the coefficients for the sines.

\[\begin = \frac\int_^ & = \frac\left[ \right]\\ & = \frac\left[ \right]\end\] As with the coefficients for the cosines will probably be easier to do each of these individually. \right|_^0 = \frac\left( \right) = \frac\left( \right)\] \[\begin\int_^ & = \left.

\right|_0^L\ & = \left( \right)\left( \right)\ & = \left( \right)\left( \right) = - \frac\end\] So, if we put all of this together we have, \[\begin = \frac\int_^ & = \frac\left[ \right]\ & = \frac\left[ \right] = - \frac\left( \right)\hspace\hspacen = 1,2,3, \ldots \end\] So, after all that work the Fourier series is, \[\beginf\left( x \right) & = \sum\limits_^\infty \sum\limits_^\infty \ & = \sum\limits_^\infty \sum\limits_^\infty \ & = L \sum\limits_^\infty - \sum\limits_^\infty \end\] As we saw in the previous example there is often quite a bit of work involved in computing the integrals involved here.

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Doing this gives, \[\int_^ = \sum\limits_^\infty \sum\limits_^\infty \] We can now take advantage of the fact that the sines and cosines are mutually orthogonal.

The integral in the second series will always be zero and in the first series the integral will be zero if \(n \ne m\) and so this reduces to, \[\int_^ = \left\{ \right.\] Solving for \(\) gives, \[\begin& = \frac\int_^\ & = \frac\int_^\hspace\hspacem = 1,2,3, \ldots \end\] Now, do it all over again only this time multiply both sides by \(\sin \left( \right)\), integrate both sides from –\(L\) to \(L\) and interchange the integral and summation to get, \[\int_^ = \sum\limits_^\infty \sum\limits_^\infty \] In this case the integral in the first series will always be zero and the second will be zero if \(n \ne m\) and so we get, \[\int_^ = \left( L \right)\] Finally, solving for \(\) gives, \[ = \frac\int_^\hspace\hspacem = 1,2,3, \ldots \] In the previous two sections we also took advantage of the fact that the integrand was even to give a second form of the coefficients in terms of an integral from 0 to \(L\).

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